\(\begin{aligned}
I & =\int_0^{\pi / 4} \frac{d x}{\cos ^3 x \sqrt{2 \sin 2 x}}=\int_0^{\pi / 4} \frac{d x}{\cos ^3 x \sqrt{4 \sin x \cos x}} \\
& =\int_0^{\pi / 4} \frac{d x}{2 \cos ^4 x \sqrt{\tan x}}=\frac{1}{2} \int_0^{\pi / 4} \frac{\sec ^2 x\left(1+\tan ^2 x\right)}{\sqrt{\tan x}} d x
\end{aligned}\)
Let, \(\tan x=t^2\), then at \(x=0, t=0\) and at
\(x=\frac{\pi}{4}, t=1 \text { and } \sec ^2 x d x=2 t \mathrm{dt}\)
\(\begin{aligned}
\text{So, } I & =\int_0^1 \frac{\left(1+t^4\right) t d t}{t}=\int_0^1\left(1+t^4\right) d t \\
& =\left[t+\frac{t^5}{5}\right]_0^1=1+\frac{1}{5}=\frac{6}{5}
\end{aligned}\)